Question

A fair coin is tossed a fixed number of times. If the probability of getting $7$ heads is equal to probability of getting $9$ heads, then the probability of getting $2$ heads is :

• A. $\frac{15}{2^{12}}$
• B. $\frac{15}{2^{13}}$
• C. $\frac{15}{2^{14}}$
• D. $\frac{15}{2^8}$

Answer 1

The correct option is B $\frac{15}{2^{13}}$

Let $X$ denote the number of heads.
$P(X=9)=P(X=7)$
$\Rightarrow {}^n{C}_9{\left(\frac{1}{2}\right)}^{n-9}\times {\left(\frac{1}{2}\right)}^9={}^n{C}_7{\left(\frac{1}{2}\right)}^{n-7}\times {\left(\frac{1}{2}\right)}^7$
$\Rightarrow {}^n{C}_9{\left(\frac{1}{2}\right)}^n={}^n{C}_7{\left(\frac{1}{2}\right)}^n$
$\Rightarrow n=9+7=16$

$\therefore P(X=2)={}^{16}{C}_2\times {\left(\frac{1}{2}\right)}^{14}\times {\left(\frac{1}{2}\right)}^2$
$={}^{16}{C}_2\times {\left(\frac{1}{16}\right)}^{16}=\frac{15}{2^{13}}$